MATH 332 - ADVANCED CALCULUS
- Green’s Theorem
Question Bank - Set 2
Liberty University
Question 1
Question
Let Cbe the curve given by x2+y2= 1 oriented counterclockwise. Calculate
the line integral HC(x2y2−y2)dx + (x3−xy)dy using Green’s Theorem.
Solution
Step 1: Verify that the curve Cencloses a simply connected region D. In this
case, Dis the unit disk centered at the origin.
Step 2: Write the given line integral as a double integral over Dusing Green’s
Theorem. Green’s Theorem states:
IC
P dx +Q dy =ZZD∂Q
∂x −∂P
∂y dA
where P=x2y2−y2and Q=x3−xy. Then, the double integral becomes:
ZZD∂Q
∂x −∂P
∂y dA =ZZD∂
∂x(x3−xy)−∂
∂y (x2y2−y2)dA
Step 3: Compute the partial derivatives:
∂Q
∂x =∂
∂x(x3−xy) = 3x2−y
∂P
∂y =∂
∂y (x2y2−y2) = 2x2y−2y
Step 4: Substitute the partial derivatives back into the double integral:
ZZD(3x2−y)−(2x2y−2y)dA =ZZDx2−2x2y+ydA
Step 5: Convert the double integral to polar coordinates since Dis a disk:
ZZDx2−2x2y+ydA =Z2π
0Z1
0(rcos θ)2−2(rcos θ)2(rsin θ)+(rsin θ)r drdθ
Step 6: Evaluate the double integral:
=Z2π
0Z1
0r2cos2θ−2r3cos2θsin θ+rsin θr drdθ
=Z2π
0Z1
0r3(cos2θ−2rcos2θsin θ+ sin θ)drdθ
=Z2π
0r4
4(cos2θ−2rcos2θsin θ+ sin θ)1
0
dθ
=Z2π
01
4(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
4Z2π
0
(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
41
2(θ+ sin θcos θ−2 sin2θ)2π
0
=1
41
2(2π+ 0 −0)
Question 2
Question
Let Cbe the curve defined by x=t3and y=t2for 0≤t≤1. Use Green’s
Theorem to evaluate HC(x2+y2)dx +xy dy.
Solution
Step 1: We start by parameterizing the curve Cusing x=t3and y=t2. Then
dx = 3t2dt and dy = 2t dt. The curve Cbecomes {(t3, t2) : 0 ≤t≤1}.
Step 2: We rewrite the line integral HC(x2+y2)dx +xy dy using the param-
eterization:
IC
(x2+y2)dx +xy dy =Z1
0
((t3)2+ (t2)2)·3t2dt + (t3·t2)·2t dt
Step 3: Simplify the integrand:
Z1
0
(t6+t4)·3t2dt + 2t4dt =Z1
0
(3t8+ 3t6)dt + 2t4dt
2
Step 4: Integrate the terms one by one:
Z1
0
3t8dt +Z1
0
3t6dt +Z1
0
2t4dt
Step 5: Calculate the integrals:
3
9t91
0
+3
7t71
0
+2
5t51
0
Step 6: Substitute the upper and lower limits and simplify the expression:
3
9+3
7+2
5=1
3+3
7+2
5=35 + 45 + 42
105 =122
105
Step 7: Therefore, the value of the line integral is 122
105 .
Question 3
Question
Let Cbe the curve formed by the intersection of the plane z=x+yand
the cylinder x2+y2= 1, oriented counterclockwise when viewed from above.
Use Green’s Theorem to evaluate the line integral HCF·drwhere F(x, y, z) =
⟨x2, y2, z2⟩.
Solution
To evaluate the line integral using Green’s Theorem, we first need to find the
curve Cthat bounds the region D.
Step 1: Find the Region D
The curve Cis the intersection of the plane z=x+yand the cylinder x2+y2= 1.
Solving for zin terms of xand y, we have z=x+y. This gives us the region
Dbounded by the circle x2+y2= 1 in the xy-plane.
Step 2: Calculate the Line Integral Using Green’s Theorem
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
where F(x, y, z) = ⟨P, Q, R⟩. In this case, F(x, y, z) = ⟨x2, y2, z2⟩, so P=x2,
Q=y2, and R=z2.
The partial derivatives are: ∂Q
∂x = 0
∂P
∂y = 0
3
Thus, the line integral simplifies to:
IC
F·dr=ZZD
(0 −0)dA = 0
Therefore, the value of the line integral HCF·dris 0.
Question 4
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using Green’s Theorem.
Solution
Step 1: Determine the region Denclosed by the curve C. The curve Cis a
circle centered at the origin with radius 2. Therefore, the region Denclosed by
Cis the interior of the circle.
Step 2: Check if Green’s Theorem applies. Green’s Theorem states that for
a simple, closed, positively oriented curve Cenclosing a region D, if P(x, y)
and Q(x, y)have continuous partial derivatives on an open region containing D,
then IC
P dx +Qdy =ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Verify the conditions for Green’s Theorem. Given P(x, y) = 3x2y−y3
and Q(x, y) = x3−2xy2,
∂Q
∂x = 3x2−2y2and ∂P
∂y = 3x2−3y2.
Both ∂Q
∂x and ∂P
∂y are continuous on D, so Green’s Theorem applies.
Step 4: Apply Green’s Theorem. The line integral can be written as
IC
(3x2y−y3)dx + (x3−2xy2)dy =ZZD∂(x3−2xy2)
∂x −∂(3x2y−y3)
∂y dA.
Simplifying the right-hand side gives
ZZD
(3x2−2y2−3x2+ 3y2)dA =ZZD
y2dA.
Step 5: Calculate the double integral. Since the region Dis a circle of radius
2 centered at the origin, we can convert the integral to polar coordinates with
0≤r≤2and 0≤θ≤2π. Therefore, the double integral becomes
Z2π
0Z2
0
r2·rdrdθ.
4
Step 6: Evaluate the integral. Computing the double integral gives
Z2π
0Z2
0
r3drdθ =Z2π
0r4
42
0
dθ =Z2π
0
4dθ = 8π.
Hence, the value of the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using
Green’s Theorem is 8π.
Question 5
Question
Let Cbe the curve formed by the intersection of the plane 2x+y+z= 4 and
the cylinder x2+y2= 1. Use Green’s Theorem to evaluate the line integral
HC(y2+z2)dx + (z2+x2)dy + (x2+y2)dz.
Solution
To use Green’s Theorem, we first need to find a vector field Fsuch that ∇×F=
y2+z2i+z2+x2j+x2+y2k.
Let’s parameterize the curve C. We can use cylindrical coordinates to de-
scribe Cas x= cos(t), y = sin(t), z = 4 −2 cos(t)−sin(t)where 0≤t≤2π.
Step 1: Calculate the partial derivatives of x,y, and zwith respect to t.
dx
dt =−sin(t),dy
dt = cos(t),dz
dt = 2 sin(t)−cos(t)
Step 2: Calculate the line integral. Using Green’s Theorem, we have
IC
F·dr=ZZD
(∇ × F)·ndA
Where Dis the region enclosed by curve Cand nis the outward unit normal
to D.
Now we need to find the unit normal vector n=1
√6(−1,−2,1).
Step 3: Calculate the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2+z2z2+x2x2+y2
= (2y−2y, 2z−2z, 2x−2x) = 0
Since the curl is zero, the line integral over Cis 0.
Question 6
Question
Let Cbe the curve consisting of the line segment from (0,0) to (2,2) followed by
the line segment from (2,2) to (4,0) traversed in the counterclockwise direction.
5
Let Dbe the region enclosed by C. Use Green’s Theorem to evaluate the line
integral HCF·dr, where F(x, y) = (y2,sin(xy)).
Solution
Using Green’s Theorem, we have that
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = (P, Q).
Step 1: Compute the partial derivatives of Pand Q
∂Q
∂x =∂
∂x(sin(xy)) = ycos(xy),
∂P
∂y =∂
∂y (y2) = 2y.
Step 2: Evaluate the double integral over region DUsing the given curve C,
we can parameterize the line segments as follows: For the first segment: x=t,
y=t,0≤t≤2. For the second segment: x= 2t+ 2,y=−2t+ 2,0≤t≤1.
Therefore, the integral becomes:
ZZD
(ycos(xy)−2y)dA
=Z2
0Zt
0
(tcost2−2t)dy dx +Z1
0Z−2t+2
2t+2
((−2t+ 2) cos(−2t(2t+ 2)) −2(−2t+ 2)) dy dx
=Z2
0 t2sint2
2−t2!dx +Z1
0
(2t(2 cos(2t(2t+ 2)) −2) −2(2t−2)) dx.
This integral may be simplified further, but the final answer depends on the
specific values obtained after solving the nested integrals.
Question 7
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(2x−y)dx + (x+y)dy using Green’s Theorem.
Solution
1. We first need to determine the region enclosed by the curve C, which is a
circle centered at the origin with radius 2. Let Dbe the region enclosed
by C.
6
2. Green’s Theorem states that for a simply connected region Dbounded by a
positively-oriented, piecewise-smooth, simple closed curve C, if
F(x, y) =
(P(x, y), Q(x, y)) is a continuously differentiable vector field defined on an
open region containing D, then
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where dr =dxi+dyjand dA =dxdy.
3. In our case,
F(x, y) = (2x−y, x+y),P(x, y) = 2x−y, and Q(x, y) = x+y.
We can rewrite the line integral as
IC
(2x−y)dx + (x+y)dy =ZZD
(2 −1) dA =ZZD
dA.
4. To evaluate RRDdA, we convert the double integral into polar coordinates.
Let x=rcos θand y=rsin θ. Then dA =r dr dθ, and the integral
becomes
ZZD
dA =Z2π
0Z2
0
r dr dθ.
5. Solving the inner integral, we have
Z2
0
r dr =r2
2
2
0= 4.
6. Substituting this back into the double integral, we get
Z2π
0Z2
0
r dr dθ =Z2π
0
4dθ = 8π.
7. Therefore, the line integral HC(2x−y)dx + (x+y)dy is equal to 8π.
Question 8
Question
Let Cbe the curve given by the intersection of the surfaces z=x2and y=x3,
oriented counterclockwise when viewed from above. Use Green’s Theorem to
evaluate the line integral HC(x2+y2)dx + 2xydy.
Solution
Step 1: Determine the region Denclosed by the curve C. To find the region
Denclosed by C, we need to find the intersection points of the surfaces z=x2
and y=x3. Solving the equations z=x2and y=x3simultaneously, we get
7
z=x2and z=y2/3. So Dlies between the curves y=x3and y=x2/3. Now,
we can express the boundaries of Das x=y1/3and x=y2/3.
Step 2: Calculate the partial derivatives. Let P(x, y) = x2+y2and
Q(x, y) = 2xy. We need to compute ∂Q
∂x −∂P
∂y .
∂Q
∂x = 2yand ∂P
∂y = 2y
Step 3: Apply Green’s Theorem. By Green’s Theorem,
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA
=ZZD
(2y−2y)dA
=ZZD
0dA
= 0
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
curve Cis 0.
Question 9
Question
Let Cbe the curve defined by C(t) = (sin t, cos t)for 0≤t≤π/2. Calculate
the line integral HCF·dr, where F(x, y) = (−y, x), using Green’s Theorem.
Solution
Step 1: Find the region Denclosed by the curve C.
• We have C(t) = (sin t, cos t)for 0≤t≤π/2which represents the quarter-
circle of radius 1 centered at the origin in the first quadrant.
• Therefore, the region Denclosed by Cis the area of this quarter-circle.
Step 2: Verify if the hypotheses of Green’s Theorem are satisfied.
• Green’s Theorem states that for a positively oriented simple closed curve
C, the line integral of Falong Cis equal to the double integral of (∂Q/∂x−
∂P/∂y)over the region Denclosed by C, where F= (P, Q).
• Calculate ∂Q
∂x −∂P
∂y for F(x, y) = (−y, x):
∂Q
∂x −∂P
∂y = 1 −(−1) = 2
This is a nonzero constant, which is good.
8
Step 3: Apply Green’s Theorem to compute the line integral using the double
integral.
• Green’s Theorem gives
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
=ZZD
2dA
= 2 ZZD
dA
= 2 ·(Area of D)
Since the area of the quarter-circle is 1
4π(1)2=π
4, the line integral is
2·π
4=π
2.
Thus, the line integral HCF·dr=π
2.
Question 10
Question
Let Cbe the closed curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral HC(y2−x)dx +x2dy.
Solution
Step 1: Let’s first find the region Denclosed by the curve C, which is the circle
with radius 2 centered at the origin. We can see that Dis the disk x2+y2≤4.
Step 2: Green’s Theorem states that for a positively oriented simple closed
curve Cand a region Dwhose boundary is C, if P(x, y)and Q(x, y)have
continuous partial derivatives on an open region containing D, then
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Let P(x, y) = y2−xand Q(x, y) = x2. Then, we have
∂Q
∂x = 2xand ∂P
∂y = 2y.
Step 4: We can rewrite the line integral as
IC
(y2−x)dx +x2dy =ZZD
(2x−2y)dA.
9
Step 5: Since Dis the disk of radius 2 centered at the origin, we can switch
to polar coordinates by letting x=rcos(θ)and y=rsin(θ). Then dA =rdrdθ,
and the integral becomes
Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ.
Step 6: This simplifies to
Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ =Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=16
3[sin(θ) + cos(θ)]2π
0
=16
3[sin(2π) + cos(2π)−(sin(0) + cos(0))]
=16
3(0 + 1 −(0 + 1))
= 0 .
Therefore, the value of the line integral HC(y2−x)dx +x2dy is 0.
Question 11
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise, and let D
be the region enclosed by C. Use Green’s Theorem to evaluate the line integral
HC(x2+y2)dx +xydy.
Solution
Step 1: First, let’s calculate the partial derivatives of N(x, y) = x2+y2and
M(x, y) = xy.
Given M(x, y) = xy, we have:
∂M
∂y =x
Given N(x, y) = x2+y2, we have:
∂N
∂x = 2x
Step 2: Now, apply Green’s Theorem, which states:
IC
(Mdx +N dy) = ZZD∂N
∂x −∂M
∂y dA
10
So, we have:
IC
(x2+y2)dx +xydy =ZZD
(2x−x)dA
Step 3: Simplify the double integral:
ZZD
(2x−x)dA =ZZD
xdA
Step 4: The region Denclosed by the curve Cis a circle with radius 2. Thus,
we can convert the double integral to polar coordinates:
ZZD
xdA =Z2π
0Z2
0
(rcos θ)rdrdθ
Step 5: Evaluate the integral in polar coordinates:
Z2π
0Z2
0
(rcos θ)rdrdθ =Z2π
023
3cos θdθ =16
3π
Therefore, the value of the line integral HC(x2+y2)dx +xydy is 16
3π.
Question 12
Question
Let Cbe the curve given by the intersection of the plane z= 4 −xand the
cylinder x2+y2= 4. Use Green’s Theorem to find the circulation of the vector
field F= (y2, z, x)around the curve C.
Solution
To apply Green’s Theorem, we need to find the normal vector nto the curve C.
Step 1: Find the parametric equations for the curve C.
The intersection of the plane z= 4 −xand the cylinder x2+y2= 4 can be
parameterized as follows:
x= 2 cos(t), y = 2 sin(t), z = 4 −2 cos(t)
for 0≤t≤2π.
Step 2: Calculate the circulation of the vector field F.
The circulation of Faround Cis given by the line integral:
IC
F·dr=ZC
F·Tds
where Tis the unit tangent vector to the curve C.
11
The unit tangent vector Tis given by dr
dt . Hence, we have:
T= (−2 sin(t),2 cos(t),2 sin(t))
Now, we can compute the line integral:
IC
F·dr=Z2π
0
F(2 sin(t),4−2 cos(t),2 cos(t)) ·(−2 sin(t),2 cos(t),2 sin(t)) dt
Calculating this line integral will give us the circulation of the vector field F
around the curve C.
Question 13
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(x2y−xy2)dx + (x3y−y3)dy using Green’s Theorem.
Solution
Step 1: Determine the region enclosed by the curve C. We can see that Cis a
circle of radius 2centered at the origin.
Step 2: Write the given line integral in terms of a double integral by apply-
ing Green’s Theorem. Green’s Theorem states that for a vector field F= (P, Q),
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by the curve C.
In this case, F= (x2y−xy2, x3y−y3), and we have:
P=x2y−xy2,
Q=x3y−y3.
Calculating the partial derivatives:
∂Q
∂x = 3x2y,
∂P
∂y =x2−2xy.
So, the line integral becomes:
IC
(x2y−xy2)dx + (x3y−y3)dy =ZZD
(3x2y−x2+ 2xy)dA.
12
Step 3: Find the integral over the region D. Since Dis a disk of radius 2,
we can express Din polar coordinates as 0≤r≤2,0≤θ≤2π. Therefore, we
have:
ZZD
(3x2y−x2+ 2xy)dA =Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ.
Step 4: Evaluate the double integral. Let’s simplify the integrand:
3r3sin2(θ)−r2+ 2rsin(θ) = 3r4sin2(θ)−r2+ 2rsin(θ).
Then, compute the double integral:
Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ
=Z2π
03
5r5sin2(θ)−1
3r3+r2sin(θ)
2
0
dθ
=Z2π
096
5sin2(θ)−8
3+ 4 sin(θ)dθ
=96
5θ
2−sin(2θ)
4−8
3θ−4 cos(θ)
2π
0
= 0.
Step 5: Conclusion. Therefore, the line integral HC(x2y−xy2)dx + (x3y−
y3)dy around the curve Cis equal to 0.
Question 14
Question
Let Cbe the curve consisting of the line segment from (1,0) to (0,1), followed
by the arc of the circle x2+y2= 1 from (0,1) to (1,0). Use Green’s Theorem
to evaluate the line integral HC(x2−y)dx +xydy.
Solution
Step 1: First, we need to parameterize the curve C. Let’s denote the line
segment as C1and the arc of the circle as C2. We can parameterize C1as
r1(t) = (1 −t, t)for 0≤t≤1, and C2as r2(t) = (cos(t),sin(t)) for 0≤t≤π.
Step 2: Next, we need to compute the partial derivatives of P(x, y) = x2−y
and Q(x, y) = xy. We have:
∂Q
∂x =yand ∂P
∂y =−1
13
Step 3: Now, we can apply Green’s Theorem, which states that for a counterclockwise-
oriented simple closed curve Cand a region Dbounded by C, we have:
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA
Step 4: Substituting the given Pand Q, and the computed partial deriva-
tives, into Green’s Theorem, we get:
IC
(x2−y)dx +xydy =ZZD
(y+ 1)dA
Step 5: Since Dis the unit disk, we can rewrite the double integral in polar
coordinates. Thus, the line integral becomes:
IC
(x2−y)dx +xydy =Z2π
0Z1
0
(rsin(θ) + 1)rdrdθ
Step 6: Solving the double integral, we obtain:
=Z2π
0Z1
0
r2sin(θ) + rdrdθ
=Z2π
01
2sin(θ) + 1
2dθ
=−1
2cos(θ) + θ
22π
0
=π
Thus, the value of the line integral HC(x2−y)dx +xydy around the curve C
is π.
Question 15
Question
Let Cbe the square with vertices at (0,0),(2,0),(2,2), and (0,2). Evaluate
the line integral HC(x2y−y2)dx + (x2+xy)dy using Green’s theorem.
Solution
Step 1: Calculate the partial derivatives of the given vector field F(x, y) =
(x2y−y2, x2+xy).
∂
∂y (x2y−y2) = x2−2y
∂
∂x(x2+xy) = 2x+y
14
Step 2: Compute the double integrals over the region Denclosed by the
square C.
ZZD∂Q
∂x −∂P
∂y dA
ZZD(2x+y)−(x2−2y)dA
Step 3: Find the limits of integration for the double integral. The square C
can be parameterized by x=t, y =tfor 0≤t≤2.
Step 4: Evaluate the double integral.
Z2
0Z2
0
((2x+y)−(x2−2y))dydx
Z2
0Z2
0
(2t+t−t2+ 2t)dydx
Z2
0Z2
0
(5t−t2)dydx
Z2
0
[(5t−t2)y]2
0dx
Z2
0
(10 −4t)dx
Z2
0
10dx −Z2
0
4tdx
10(2) −422
2−02
2
20 −4(2)
20 −8 = 12
Therefore, the value of the line integral HC(x2y−y2)dx + (x2+xy)dy using
Green’s theorem is 12 .
Question 16
Question
Let Cbe the circle given by the curve parameterization r(t) = (3 cos t, 3 sin t)
where 0≤t≤2π. Calculate the circulation HC
f·dr for the vector field
f(x, y) = (−y, x)using Green’s Theorem.
15
Solution
Step 1: Calculate the circulation directly:
IC
f·dr =Z2π
0
f(r(t)) ·r′(t)dt
=Z2π
0
(−3 sin t, 3 cos t)·(−3 sin t, 3 cos t)dt
=Z2π
0
9 sin2t+ 9 cos2tdt
=Z2π
0
9dt
= 18π
Step 2: Use Green’s Theorem to calculate the circulation: Green’s Theorem
states that for a region Renclosed by a simple closed curve C, and a vector
field
F=Pi+Qjdefined on an open region containing R:
IC
F·dr =ZZR∂Q
∂x −∂P
∂y dA
Step 3: Validate Green’s Theorem for our vector field
f(x, y) = (−y, x): Let
P(x, y) = −yand Q(x, y) = x, then:
∂Q
∂x = 1,∂P
∂y =−1
∂Q
∂x −∂P
∂y = 2
Step 4: Green’s Theorem gives:
IC
f·dr =ZZR
2dA
= 2 ·Area(R)
Step 5: The circle Cencloses the region Rcorresponding to the disk x2+y2≤
9. The area of this disk is π·32= 9π. Therefore:
IC
f·dr = 2 ·9π= 18π
Thus, the circulation of the vector field
f(x, y)=(−y, x)around the circle
Cis 18π.
16
Question 17
Question
Let Cbe the curve given by x2+y2= 4 oriented counterclockwise, and let D
be the region bounded by C. Use Green’s Theorem to evaluate the line integral
IC
(x2y−y2)dx + (x2+xy)dy.
Solution
Step 1: Identify the region Dbounded by the curve C.The curve Cis
a circle centered at the origin with radius 2. Therefore, the region Dbounded
by Cis the interior of the circle.
Step 2: Apply Green’s Theorem. Green’s Theorem states that for
a region Dbounded by a positively oriented, piecewise-smooth, simple closed
curve C, and for a vector field F(x, y)=(P(x, y), Q(x, y)) with continuous
partial derivatives on an open region containing D, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where dr= (dx, dy)is the differential path element and dA =dx dy is the area
element.
In our case, P(x, y) = x2y−y2and Q(x, y) = x2+xy. We want to calculate
the line integral HCF·dr, which is equal to the double integral RRD∂Q
∂x −∂P
∂y dA.
Step 3: Calculate the double integral. We first need to find ∂Q
∂x and
∂P
∂y :
∂Q
∂x = 2x+yand ∂P
∂y =x2−2y.
Therefore, the expression ∂Q
∂x −∂P
∂y = 2x+y−x2+ 2y.
Step 4: Evaluate the double integral over D.Now, we can proceed to
evaluate the double integral over the region D:
ZZD
(2x+y−x2+ 2y)dA =ZZD
(2x−x2+ 3y)dA.
To simplify this integral, we can convert to polar coordinates (x=rcos θ,
y=rsin θ) with dA =r dr dθ:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ.
Step 5: Compute the integral. Evaluating the integral, we get:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ =π.
So, the value of the line integral HC(x2y−y2)dx+(x2+xy)dy over the curve
Cis π.
17
Question 18
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise. Using
Green’s Theorem, evaluate the line integral HC
y
x2+y2dx +x
x2+y2dy.
Solution
Step 1: Determine the region enclosed by the curve C. The curve Cis a circle
with radius 2 centered at the origin. The region enclosed by Cis the interior of
this circle.
Step 2: Express the line integral as a double integral using Green’s Theorem.
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = y
x2+y2,x
x2+y2and Dis the region enclosed by C.
In this case, P(x, y) = y
x2+y2and Q(x, y) = x
x2+y2.
Step 3: Compute the partial derivatives.
∂Q
∂x =(x2+y2)−x(2x)
(x2+y2)2=y2−x2
(x2+y2)2,
∂P
∂y =(x2+y2)−y(2y)
(x2+y2)2=x2−y2
(x2+y2)2.
Step 4: Evaluate the double integral.
ZZDy2−x2
(x2+y2)2−x2−y2
(x2+y2)2dA =ZZD
0dA = 0.
Therefore, the value of the line integral HC
y
x2+y2dx+x
x2+y2dy using Green’s
Theorem is 0.
Question 19
Question
Let Cbe the circle centered at the origin with radius 2 oriented counterclockwise,
and let Dbe the region enclosed by C. Given the vector field F(x, y)=(x2+
y2,−x), calculate the line integral of Faround the boundary of Dusing Green’s
Theorem.
18
Solution
Step 1: Compute ∂Q
∂x and ∂P
∂y .
∂Q
∂x =∂
∂x(−x) = −1
∂P
∂y =∂
∂y (x2+y2) = 2y
Step 2: Apply Green’s Theorem, which states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
Step 3: Evaluate the double integral over D.
ZZD
(−1−2y)dA =−ZZD
(1 + 2y)dA
Step 4: Convert to polar coordinates since Cis a circle. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =rdrdθ, with rranging from 0 to 2 and θ
ranging from 0 to 2π.
Step 5: Replace xand yin the integral with rcos(θ)and rsin(θ).
−Z2π
0Z2
0
(1 + 2rsin(θ))rdrdθ
Step 6: Evaluate the integral.
−Z2π
0Z2
0
r+ 2r2sin(θ)drdθ
−Z2π
0 r2
2+2r3
3sin(θ)
2
0!dθ
−Z2π
02 + 16
3sin(θ)dθ
−2θ−16
3cos(θ)
2π
0
−(4π−0) = −4π
Therefore, the line integral of Faround the boundary of Dis −4π.
Question 20
Question
Let Cbe the curve given by the intersection of the cylinder x2+y2= 1 and
the plane 2x+y+z= 3. Use Green’s Theorem to evaluate the line integral
HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).
19
Solution
To apply Green’s Theorem, we first need to find a region Din the xy-plane
that contains the curve Cas its boundary. Since Cis the intersection of the
cylinder x2+y2= 1 and the plane 2x+y+z= 3, we can rewrite zin terms
of xand yas z= 3 −2x−y. So the projection of Conto the xy-plane is the
circle x2+y2= 1.
Let Dbe the disk enclosed by the circle x2+y2= 1. Then we have ∂D =C.
By Green’s Theorem, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z)) = (y+z, x +z, x +y).
Now, we compute:
∂Q
∂x = 1,
∂P
∂y = 1,
∂Q
∂x −∂P
∂y = 0.
Therefore, the line integral can be simplified as:
IC
F·dr=ZZD
0dA = 0.
So, the value of the line integral is 0.
Question 21
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Let
F(x, y)=(x2+y2)i+x2j. Calculate the line integral HCF·dr using Green’s
Theorem.
Solution
Step 1: Find the region Denclosed by the curve C. Since x2+y2= 4 represents
a circle of radius 2 centered at the origin, the region Denclosed by Cis also
the region inside this circle.
Step 2: Calculate ∂Q
∂x and ∂P
∂y . Given F(x, y)=(x2+y2)i+x2j, we have
P(x, y) = x2+y2and Q(x, y) = x2.
Calculating the partial derivatives:
∂Q
∂x = 2xand ∂P
∂y = 2y
20
Step 3: Apply Green’s Theorem. Green’s Theorem states: HCF·dr =
RRD∂Q
∂x −∂P
∂y dA
Substitute the partial derivatives:
IC
F·dr =ZZD
(2x−2y)dA
Step 4: Convert the integral to polar coordinates. In polar coordinates,
dA =rdrdθ and x=rcos(θ), y =rsin(θ).
The integral becomes:
IC
F·dr =Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ
Step 5: Evaluate the integral.
IC
F·dr =Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ
=Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=−16
3sin(θ)−16
3cos(θ)2π
0
=−16
3(sin(2π) + cos(2π)−sin(0) −cos(0))
=−16
3(0 + 1 −0−1) = 32
3
Therefore, the line integral HCF·dr =32
3.
Question 22
Question
Let Cbe the curve defined by x(t) = 3 cos(t),y(t) = 2 sin(t)for 0≤t≤2π.
Evaluate the line integral HC(x2+y2)dx +xy dy.
21
Solution
Step 1: Calculate the derivatives of x(t)and y(t)with respect to t.
dx
dt =−3 sin(t)and dy
dt = 2 cos(t)
Step 2: Compute dx and dy.
dx =dx
dt dt =−3 sin(t)dt and dy =dy
dt dt = 2 cos(t)dt
Step 3: Substitute x(t),y(t),dx, and dy into the line integral.
IC
(x2+y2)dx+xy dy =Z2π
0
((3 cos(t))2+(2 sin(t))2)(−3 sin(t))+3 cos(t)·2 cos(t)dt
Step 4: Simplify the integrand.
Z2π
0
(9 cos2(t)+4 sin2(t))(−3 sin(t))+6 cos(t) sin(t)dt =Z2π
0
(−27 cos2(t) sin(t)−12 sin3(t)+6 cos(t) sin(t)) dt
Step 5: Evaluate the integral.
=−27 Z2π
0
cos2(t) sin(t)dt −12 Z2π
0
sin3(t)dt + 6 Z2π
0
cos(t) sin(t)dt
Step 6: Use trigonometric identities to simplify and evaluate each integral.
=−27 −cos3(t)
32π
0
−12 −cos2(t)
2+cos4(t)
42π
0
+ 6 −cos2(t)
22π
0
Step 7: Substitute 2πand 0into the expressions and simplify to find the
final answer.
Question 23
Question
Let Cbe the positively oriented circle given by x2+y2= 4. Evaluate the line
integral HC(y3dx +x3dy)using Green’s Thejsonreorem.
Solution
Step 1: We need to express the line integral as a double integral over the region
enclosed by Cusing Green’s Theorem. Green’s Theorem relates the line integral
of a vector field over a closed curve to the double integral of the divergence of a
related vector field over the region enclosed by the curve. The theorem states:
IC
(P dx +Qdy) = ZZR∂Q
∂x −∂P
∂y dA
22
Step 2: The given line integral HC(y3dx +x3dy)can be written as HCx3dy +
y3dx. Let P(x, y) = x3and Q(x, y) = y3. Then, ∂Q
∂x = 0 and ∂P
∂y = 0. Hence,
the double integral reduces to zero.
Step 3: Since the double integral over the region enclosed by Cis zero, the
value of the line integral HC(y3dx +x3dy)is also zero.
Therefore, the value of the line integral HC(y3dx +x3dy)over the circle
x2+y2= 4 is 0.
Question 24
Question
Evaluate the line integral HC(x2+y2)dx+2xydy, where Cis the circle x2+y2= 4,
oriented counterclockwise.
Solution
Step 1: We will first rewrite the given line integral in terms of a potential
function. Let P(x, y) = x2+y2and Q(x, y) = 2xy. Then, we have ∂Q
∂x = 2y=
∂P
∂y , which implies that the integrand is exact. By Green’s Theorem, we have
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by C, which is the circle x2+y2= 4.
Step 2: The region Dcan be described using the polar coordinates x=
rcos(θ),y=rsin(θ), with 0≤r≤2and 0≤θ≤2π. Thus, dA =rdrdθ. The
integral becomes
ZZD∂Q
∂x −∂P
∂y dA =Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ.
Step 3: We can now evaluate the double integral:
Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ =Z2π
0Z2
0
2r2sin(θ)−2r2drdθ
=Z2π
0 2
3r3sin(θ)−r3r=2
r=0!dθ
=Z2π
016
3sin(θ)−8dθ
=−16
3cos(θ)−8θ2π
0
=−16
3(cos(2π)−cos(0)) −8(2π−0)
=−16
3−16π.
23
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
circle x2+y2= 4 oriented counterclockwise is −16
3−16π.
Question 25
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral
IC
(x2−y2)dx + 2xy dy.
Solution
Step 1: First, let’s calculate the partial derivatives of the given vector field
F(x, y) = (x2−y2,2xy).
∂
∂y (x2−y2) = −2y, ∂
∂x(2xy) = 2y.
Step 2: Now, Green’s Theorem states that for a simply connected region
Dbounded by a positively oriented, piecewise-smooth, simple closed curve C,
the line integral of a vector field
F= (P, Q)along Ccan be evaluated as
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where Pand Qare the component functions of
F.
Step 3: Applying Green’s Theorem to the given line integral, we have
IC
(x2−y2)dx + 2xy dy =ZZD
(2y−(−2y)) dA = 4 ZZD
y dA.
Step 4: The region Denclosed by the curve Cis a circle of radius 2.
We can use polar coordinates to evaluate the double integral. The conversion
formula from Cartesian to polar coordinates is x=rcos θand y=rsin θ, with
dA =r dr dθ.
Step 5: Substituting y=rsin θand dA =r dr dθ into the double integral,
we get
4ZZD
y dA = 4 Z2π
0Z2
0
rsin θ·r dr dθ.
24
Step 6: Evaluating the double integral, we have
4Z2π
0Z2
0
rsin θ·r dr dθ = 4 Z2π
0r3cos θ
3
2
0sin θ dθ
= 4 Z2π
0
8
3cos θsin θ dθ
=32
3Z2π
0
1
2sin(2θ)dθ
=32
3−1
4cos(2θ)
2π
0
=32
3−1
4−(−1
4)
=32
3×0
= 0.
Therefore, the value of the line integral HC(x2−y2)dx + 2xy dy along the
curve Cis 0.
25
Step 5: Convert the double integral to polar coordinates since Dis a disk:
ZZDx2−2x2y+ydA =Z2π
0Z1
0(rcos θ)2−2(rcos θ)2(rsin θ)+(rsin θ)r drdθ
Step 6: Evaluate the double integral:
=Z2π
0Z1
0r2cos2θ−2r3cos2θsin θ+rsin θr drdθ
=Z2π
0Z1
0r3(cos2θ−2rcos2θsin θ+ sin θ)drdθ
=Z2π
0r4
4(cos2θ−2rcos2θsin θ+ sin θ)1
0
dθ
=Z2π
01
4(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
4Z2π
0
(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
41
2(θ+ sin θcos θ−2 sin2θ)2π
0
=1
41
2(2π+ 0 −0)
Question 2
Question
Let Cbe the curve defined by x=t3and y=t2for 0≤t≤1. Use Green’s
Theorem to evaluate HC(x2+y2)dx +xy dy.
Solution
Step 1: We start by parameterizing the curve Cusing x=t3and y=t2. Then
dx = 3t2dt and dy = 2t dt. The curve Cbecomes {(t3, t2) : 0 ≤t≤1}.
Step 2: We rewrite the line integral HC(x2+y2)dx +xy dy using the param-
eterization:
IC
(x2+y2)dx +xy dy =Z1
0
((t3)2+ (t2)2)·3t2dt + (t3·t2)·2t dt
Step 3: Simplify the integrand:
Z1
0
(t6+t4)·3t2dt + 2t4dt =Z1
0
(3t8+ 3t6)dt + 2t4dt
2
Step 4: Integrate the terms one by one:
Z1
0
3t8dt +Z1
0
3t6dt +Z1
0
2t4dt
Step 5: Calculate the integrals:
3
9t91
0
+3
7t71
0
+2
5t51
0
Step 6: Substitute the upper and lower limits and simplify the expression:
3
9+3
7+2
5=1
3+3
7+2
5=35 + 45 + 42
105 =122
105
Step 7: Therefore, the value of the line integral is 122
105 .
Question 3
Question
Let Cbe the curve formed by the intersection of the plane z=x+yand
the cylinder x2+y2= 1, oriented counterclockwise when viewed from above.
Use Green’s Theorem to evaluate the line integral HCF·drwhere F(x, y, z) =
⟨x2, y2, z2⟩.
Solution
To evaluate the line integral using Green’s Theorem, we first need to find the
curve Cthat bounds the region D.
Step 1: Find the Region D
The curve Cis the intersection of the plane z=x+yand the cylinder x2+y2= 1.
Solving for zin terms of xand y, we have z=x+y. This gives us the region
Dbounded by the circle x2+y2= 1 in the xy-plane.
Step 2: Calculate the Line Integral Using Green’s Theorem
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
where F(x, y, z) = ⟨P, Q, R⟩. In this case, F(x, y, z) = ⟨x2, y2, z2⟩, so P=x2,
Q=y2, and R=z2.
The partial derivatives are: ∂Q
∂x = 0
∂P
∂y = 0
3
Thus, the line integral simplifies to:
IC
F·dr=ZZD
(0 −0)dA = 0
Therefore, the value of the line integral HCF·dris 0.
Question 4
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using Green’s Theorem.
Solution
Step 1: Determine the region Denclosed by the curve C. The curve Cis a
circle centered at the origin with radius 2. Therefore, the region Denclosed by
Cis the interior of the circle.
Step 2: Check if Green’s Theorem applies. Green’s Theorem states that for
a simple, closed, positively oriented curve Cenclosing a region D, if P(x, y)
and Q(x, y)have continuous partial derivatives on an open region containing D,
then IC
P dx +Qdy =ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Verify the conditions for Green’s Theorem. Given P(x, y) = 3x2y−y3
and Q(x, y) = x3−2xy2,
∂Q
∂x = 3x2−2y2and ∂P
∂y = 3x2−3y2.
Both ∂Q
∂x and ∂P
∂y are continuous on D, so Green’s Theorem applies.
Step 4: Apply Green’s Theorem. The line integral can be written as
IC
(3x2y−y3)dx + (x3−2xy2)dy =ZZD∂(x3−2xy2)
∂x −∂(3x2y−y3)
∂y dA.
Simplifying the right-hand side gives
ZZD
(3x2−2y2−3x2+ 3y2)dA =ZZD
y2dA.
Step 5: Calculate the double integral. Since the region Dis a circle of radius
2 centered at the origin, we can convert the integral to polar coordinates with
0≤r≤2and 0≤θ≤2π. Therefore, the double integral becomes
Z2π
0Z2
0
r2·rdrdθ.
4
Step 6: Evaluate the integral. Computing the double integral gives
Z2π
0Z2
0
r3drdθ =Z2π
0r4
42
0
dθ =Z2π
0
4dθ = 8π.
Hence, the value of the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using
Green’s Theorem is 8π.
Question 5
Question
Let Cbe the curve formed by the intersection of the plane 2x+y+z= 4 and
the cylinder x2+y2= 1. Use Green’s Theorem to evaluate the line integral
HC(y2+z2)dx + (z2+x2)dy + (x2+y2)dz.
Solution
To use Green’s Theorem, we first need to find a vector field Fsuch that ∇×F=
y2+z2i+z2+x2j+x2+y2k.
Let’s parameterize the curve C. We can use cylindrical coordinates to de-
scribe Cas x= cos(t), y = sin(t), z = 4 −2 cos(t)−sin(t)where 0≤t≤2π.
Step 1: Calculate the partial derivatives of x,y, and zwith respect to t.
dx
dt =−sin(t),dy
dt = cos(t),dz
dt = 2 sin(t)−cos(t)
Step 2: Calculate the line integral. Using Green’s Theorem, we have
IC
F·dr=ZZD
(∇ × F)·ndA
Where Dis the region enclosed by curve Cand nis the outward unit normal
to D.
Now we need to find the unit normal vector n=1
√6(−1,−2,1).
Step 3: Calculate the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2+z2z2+x2x2+y2
= (2y−2y, 2z−2z, 2x−2x) = 0
Since the curl is zero, the line integral over Cis 0.
Question 6
Question
Let Cbe the curve consisting of the line segment from (0,0) to (2,2) followed by
the line segment from (2,2) to (4,0) traversed in the counterclockwise direction.
5
Let Dbe the region enclosed by C. Use Green’s Theorem to evaluate the line
integral HCF·dr, where F(x, y) = (y2,sin(xy)).
Solution
Using Green’s Theorem, we have that
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = (P, Q).
Step 1: Compute the partial derivatives of Pand Q
∂Q
∂x =∂
∂x(sin(xy)) = ycos(xy),
∂P
∂y =∂
∂y (y2) = 2y.
Step 2: Evaluate the double integral over region DUsing the given curve C,
we can parameterize the line segments as follows: For the first segment: x=t,
y=t,0≤t≤2. For the second segment: x= 2t+ 2,y=−2t+ 2,0≤t≤1.
Therefore, the integral becomes:
ZZD
(ycos(xy)−2y)dA
=Z2
0Zt
0
(tcost2−2t)dy dx +Z1
0Z−2t+2
2t+2
((−2t+ 2) cos(−2t(2t+ 2)) −2(−2t+ 2)) dy dx
=Z2
0 t2sint2
2−t2!dx +Z1
0
(2t(2 cos(2t(2t+ 2)) −2) −2(2t−2)) dx.
This integral may be simplified further, but the final answer depends on the
specific values obtained after solving the nested integrals.
Question 7
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(2x−y)dx + (x+y)dy using Green’s Theorem.
Solution
1. We first need to determine the region enclosed by the curve C, which is a
circle centered at the origin with radius 2. Let Dbe the region enclosed
by C.
6
2. Green’s Theorem states that for a simply connected region Dbounded by a
positively-oriented, piecewise-smooth, simple closed curve C, if
F(x, y) =
(P(x, y), Q(x, y)) is a continuously differentiable vector field defined on an
open region containing D, then
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where dr =dxi+dyjand dA =dxdy.
3. In our case,
F(x, y) = (2x−y, x+y),P(x, y) = 2x−y, and Q(x, y) = x+y.
We can rewrite the line integral as
IC
(2x−y)dx + (x+y)dy =ZZD
(2 −1) dA =ZZD
dA.
4. To evaluate RRDdA, we convert the double integral into polar coordinates.
Let x=rcos θand y=rsin θ. Then dA =r dr dθ, and the integral
becomes
ZZD
dA =Z2π
0Z2
0
r dr dθ.
5. Solving the inner integral, we have
Z2
0
r dr =r2
2
2
0= 4.
6. Substituting this back into the double integral, we get
Z2π
0Z2
0
r dr dθ =Z2π
0
4dθ = 8π.
7. Therefore, the line integral HC(2x−y)dx + (x+y)dy is equal to 8π.
Question 8
Question
Let Cbe the curve given by the intersection of the surfaces z=x2and y=x3,
oriented counterclockwise when viewed from above. Use Green’s Theorem to
evaluate the line integral HC(x2+y2)dx + 2xydy.
Solution
Step 1: Determine the region Denclosed by the curve C. To find the region
Denclosed by C, we need to find the intersection points of the surfaces z=x2
and y=x3. Solving the equations z=x2and y=x3simultaneously, we get
7
z=x2and z=y2/3. So Dlies between the curves y=x3and y=x2/3. Now,
we can express the boundaries of Das x=y1/3and x=y2/3.
Step 2: Calculate the partial derivatives. Let P(x, y) = x2+y2and
Q(x, y) = 2xy. We need to compute ∂Q
∂x −∂P
∂y .
∂Q
∂x = 2yand ∂P
∂y = 2y
Step 3: Apply Green’s Theorem. By Green’s Theorem,
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA
=ZZD
(2y−2y)dA
=ZZD
0dA
= 0
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
curve Cis 0.
Question 9
Question
Let Cbe the curve defined by C(t) = (sin t, cos t)for 0≤t≤π/2. Calculate
the line integral HCF·dr, where F(x, y) = (−y, x), using Green’s Theorem.
Solution
Step 1: Find the region Denclosed by the curve C.
• We have C(t) = (sin t, cos t)for 0≤t≤π/2which represents the quarter-
circle of radius 1 centered at the origin in the first quadrant.
• Therefore, the region Denclosed by Cis the area of this quarter-circle.
Step 2: Verify if the hypotheses of Green’s Theorem are satisfied.
• Green’s Theorem states that for a positively oriented simple closed curve
C, the line integral of Falong Cis equal to the double integral of (∂Q/∂x−
∂P/∂y)over the region Denclosed by C, where F= (P, Q).
• Calculate ∂Q
∂x −∂P
∂y for F(x, y) = (−y, x):
∂Q
∂x −∂P
∂y = 1 −(−1) = 2
This is a nonzero constant, which is good.
8
Step 3: Apply Green’s Theorem to compute the line integral using the double
integral.
• Green’s Theorem gives
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
=ZZD
2dA
= 2 ZZD
dA
= 2 ·(Area of D)
Since the area of the quarter-circle is 1
4π(1)2=π
4, the line integral is
2·π
4=π
2.
Thus, the line integral HCF·dr=π
2.
Question 10
Question
Let Cbe the closed curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral HC(y2−x)dx +x2dy.
Solution
Step 1: Let’s first find the region Denclosed by the curve C, which is the circle
with radius 2 centered at the origin. We can see that Dis the disk x2+y2≤4.
Step 2: Green’s Theorem states that for a positively oriented simple closed
curve Cand a region Dwhose boundary is C, if P(x, y)and Q(x, y)have
continuous partial derivatives on an open region containing D, then
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Let P(x, y) = y2−xand Q(x, y) = x2. Then, we have
∂Q
∂x = 2xand ∂P
∂y = 2y.
Step 4: We can rewrite the line integral as
IC
(y2−x)dx +x2dy =ZZD
(2x−2y)dA.
9
Step 5: Since Dis the disk of radius 2 centered at the origin, we can switch
to polar coordinates by letting x=rcos(θ)and y=rsin(θ). Then dA =rdrdθ,
and the integral becomes
Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ.
Step 6: This simplifies to
Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ =Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=16
3[sin(θ) + cos(θ)]2π
0
=16
3[sin(2π) + cos(2π)−(sin(0) + cos(0))]
=16
3(0 + 1 −(0 + 1))
= 0 .
Therefore, the value of the line integral HC(y2−x)dx +x2dy is 0.
Question 11
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise, and let D
be the region enclosed by C. Use Green’s Theorem to evaluate the line integral
HC(x2+y2)dx +xydy.
Solution
Step 1: First, let’s calculate the partial derivatives of N(x, y) = x2+y2and
M(x, y) = xy.
Given M(x, y) = xy, we have:
∂M
∂y =x
Given N(x, y) = x2+y2, we have:
∂N
∂x = 2x
Step 2: Now, apply Green’s Theorem, which states:
IC
(Mdx +N dy) = ZZD∂N
∂x −∂M
∂y dA
10
So, we have:
IC
(x2+y2)dx +xydy =ZZD
(2x−x)dA
Step 3: Simplify the double integral:
ZZD
(2x−x)dA =ZZD
xdA
Step 4: The region Denclosed by the curve Cis a circle with radius 2. Thus,
we can convert the double integral to polar coordinates:
ZZD
xdA =Z2π
0Z2
0
(rcos θ)rdrdθ
Step 5: Evaluate the integral in polar coordinates:
Z2π
0Z2
0
(rcos θ)rdrdθ =Z2π
023
3cos θdθ =16
3π
Therefore, the value of the line integral HC(x2+y2)dx +xydy is 16
3π.
Question 12
Question
Let Cbe the curve given by the intersection of the plane z= 4 −xand the
cylinder x2+y2= 4. Use Green’s Theorem to find the circulation of the vector
field F= (y2, z, x)around the curve C.
Solution
To apply Green’s Theorem, we need to find the normal vector nto the curve C.
Step 1: Find the parametric equations for the curve C.
The intersection of the plane z= 4 −xand the cylinder x2+y2= 4 can be
parameterized as follows:
x= 2 cos(t), y = 2 sin(t), z = 4 −2 cos(t)
for 0≤t≤2π.
Step 2: Calculate the circulation of the vector field F.
The circulation of Faround Cis given by the line integral:
IC
F·dr=ZC
F·Tds
where Tis the unit tangent vector to the curve C.
11
The unit tangent vector Tis given by dr
dt . Hence, we have:
T= (−2 sin(t),2 cos(t),2 sin(t))
Now, we can compute the line integral:
IC
F·dr=Z2π
0
F(2 sin(t),4−2 cos(t),2 cos(t)) ·(−2 sin(t),2 cos(t),2 sin(t)) dt
Calculating this line integral will give us the circulation of the vector field F
around the curve C.
Question 13
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(x2y−xy2)dx + (x3y−y3)dy using Green’s Theorem.
Solution
Step 1: Determine the region enclosed by the curve C. We can see that Cis a
circle of radius 2centered at the origin.
Step 2: Write the given line integral in terms of a double integral by apply-
ing Green’s Theorem. Green’s Theorem states that for a vector field F= (P, Q),
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by the curve C.
In this case, F= (x2y−xy2, x3y−y3), and we have:
P=x2y−xy2,
Q=x3y−y3.
Calculating the partial derivatives:
∂Q
∂x = 3x2y,
∂P
∂y =x2−2xy.
So, the line integral becomes:
IC
(x2y−xy2)dx + (x3y−y3)dy =ZZD
(3x2y−x2+ 2xy)dA.
12
Step 3: Find the integral over the region D. Since Dis a disk of radius 2,
we can express Din polar coordinates as 0≤r≤2,0≤θ≤2π. Therefore, we
have:
ZZD
(3x2y−x2+ 2xy)dA =Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ.
Step 4: Evaluate the double integral. Let’s simplify the integrand:
3r3sin2(θ)−r2+ 2rsin(θ) = 3r4sin2(θ)−r2+ 2rsin(θ).
Then, compute the double integral:
Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ
=Z2π
03
5r5sin2(θ)−1
3r3+r2sin(θ)
2
0
dθ
=Z2π
096
5sin2(θ)−8
3+ 4 sin(θ)dθ
=96
5θ
2−sin(2θ)
4−8
3θ−4 cos(θ)
2π
0
= 0.
Step 5: Conclusion. Therefore, the line integral HC(x2y−xy2)dx + (x3y−
y3)dy around the curve Cis equal to 0.
Question 14
Question
Let Cbe the curve consisting of the line segment from (1,0) to (0,1), followed
by the arc of the circle x2+y2= 1 from (0,1) to (1,0). Use Green’s Theorem
to evaluate the line integral HC(x2−y)dx +xydy.
Solution
Step 1: First, we need to parameterize the curve C. Let’s denote the line
segment as C1and the arc of the circle as C2. We can parameterize C1as
r1(t) = (1 −t, t)for 0≤t≤1, and C2as r2(t) = (cos(t),sin(t)) for 0≤t≤π.
Step 2: Next, we need to compute the partial derivatives of P(x, y) = x2−y
and Q(x, y) = xy. We have:
∂Q
∂x =yand ∂P
∂y =−1
13
Step 3: Now, we can apply Green’s Theorem, which states that for a counterclockwise-
oriented simple closed curve Cand a region Dbounded by C, we have:
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA
Step 4: Substituting the given Pand Q, and the computed partial deriva-
tives, into Green’s Theorem, we get:
IC
(x2−y)dx +xydy =ZZD
(y+ 1)dA
Step 5: Since Dis the unit disk, we can rewrite the double integral in polar
coordinates. Thus, the line integral becomes:
IC
(x2−y)dx +xydy =Z2π
0Z1
0
(rsin(θ) + 1)rdrdθ
Step 6: Solving the double integral, we obtain:
=Z2π
0Z1
0
r2sin(θ) + rdrdθ
=Z2π
01
2sin(θ) + 1
2dθ
=−1
2cos(θ) + θ
22π
0
=π
Thus, the value of the line integral HC(x2−y)dx +xydy around the curve C
is π.
Question 15
Question
Let Cbe the square with vertices at (0,0),(2,0),(2,2), and (0,2). Evaluate
the line integral HC(x2y−y2)dx + (x2+xy)dy using Green’s theorem.
Solution
Step 1: Calculate the partial derivatives of the given vector field F(x, y) =
(x2y−y2, x2+xy).
∂
∂y (x2y−y2) = x2−2y
∂
∂x(x2+xy) = 2x+y
14
Step 2: Compute the double integrals over the region Denclosed by the
square C.
ZZD∂Q
∂x −∂P
∂y dA
ZZD(2x+y)−(x2−2y)dA
Step 3: Find the limits of integration for the double integral. The square C
can be parameterized by x=t, y =tfor 0≤t≤2.
Step 4: Evaluate the double integral.
Z2
0Z2
0
((2x+y)−(x2−2y))dydx
Z2
0Z2
0
(2t+t−t2+ 2t)dydx
Z2
0Z2
0
(5t−t2)dydx
Z2
0
[(5t−t2)y]2
0dx
Z2
0
(10 −4t)dx
Z2
0
10dx −Z2
0
4tdx
10(2) −422
2−02
2
20 −4(2)
20 −8 = 12
Therefore, the value of the line integral HC(x2y−y2)dx + (x2+xy)dy using
Green’s theorem is 12 .
Question 16
Question
Let Cbe the circle given by the curve parameterization r(t) = (3 cos t, 3 sin t)
where 0≤t≤2π. Calculate the circulation HC
f·dr for the vector field
f(x, y) = (−y, x)using Green’s Theorem.
15
Solution
Step 1: Calculate the circulation directly:
IC
f·dr =Z2π
0
f(r(t)) ·r′(t)dt
=Z2π
0
(−3 sin t, 3 cos t)·(−3 sin t, 3 cos t)dt
=Z2π
0
9 sin2t+ 9 cos2tdt
=Z2π
0
9dt
= 18π
Step 2: Use Green’s Theorem to calculate the circulation: Green’s Theorem
states that for a region Renclosed by a simple closed curve C, and a vector
field
F=Pi+Qjdefined on an open region containing R:
IC
F·dr =ZZR∂Q
∂x −∂P
∂y dA
Step 3: Validate Green’s Theorem for our vector field
f(x, y) = (−y, x): Let
P(x, y) = −yand Q(x, y) = x, then:
∂Q
∂x = 1,∂P
∂y =−1
∂Q
∂x −∂P
∂y = 2
Step 4: Green’s Theorem gives:
IC
f·dr =ZZR
2dA
= 2 ·Area(R)
Step 5: The circle Cencloses the region Rcorresponding to the disk x2+y2≤
9. The area of this disk is π·32= 9π. Therefore:
IC
f·dr = 2 ·9π= 18π
Thus, the circulation of the vector field
f(x, y)=(−y, x)around the circle
Cis 18π.
16
Question 17
Question
Let Cbe the curve given by x2+y2= 4 oriented counterclockwise, and let D
be the region bounded by C. Use Green’s Theorem to evaluate the line integral
IC
(x2y−y2)dx + (x2+xy)dy.
Solution
Step 1: Identify the region Dbounded by the curve C.The curve Cis
a circle centered at the origin with radius 2. Therefore, the region Dbounded
by Cis the interior of the circle.
Step 2: Apply Green’s Theorem. Green’s Theorem states that for
a region Dbounded by a positively oriented, piecewise-smooth, simple closed
curve C, and for a vector field F(x, y)=(P(x, y), Q(x, y)) with continuous
partial derivatives on an open region containing D, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where dr= (dx, dy)is the differential path element and dA =dx dy is the area
element.
In our case, P(x, y) = x2y−y2and Q(x, y) = x2+xy. We want to calculate
the line integral HCF·dr, which is equal to the double integral RRD∂Q
∂x −∂P
∂y dA.
Step 3: Calculate the double integral. We first need to find ∂Q
∂x and
∂P
∂y :
∂Q
∂x = 2x+yand ∂P
∂y =x2−2y.
Therefore, the expression ∂Q
∂x −∂P
∂y = 2x+y−x2+ 2y.
Step 4: Evaluate the double integral over D.Now, we can proceed to
evaluate the double integral over the region D:
ZZD
(2x+y−x2+ 2y)dA =ZZD
(2x−x2+ 3y)dA.
To simplify this integral, we can convert to polar coordinates (x=rcos θ,
y=rsin θ) with dA =r dr dθ:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ.
Step 5: Compute the integral. Evaluating the integral, we get:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ =π.
So, the value of the line integral HC(x2y−y2)dx+(x2+xy)dy over the curve
Cis π.
17
Question 18
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise. Using
Green’s Theorem, evaluate the line integral HC
y
x2+y2dx +x
x2+y2dy.
Solution
Step 1: Determine the region enclosed by the curve C. The curve Cis a circle
with radius 2 centered at the origin. The region enclosed by Cis the interior of
this circle.
Step 2: Express the line integral as a double integral using Green’s Theorem.
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = y
x2+y2,x
x2+y2and Dis the region enclosed by C.
In this case, P(x, y) = y
x2+y2and Q(x, y) = x
x2+y2.
Step 3: Compute the partial derivatives.
∂Q
∂x =(x2+y2)−x(2x)
(x2+y2)2=y2−x2
(x2+y2)2,
∂P
∂y =(x2+y2)−y(2y)
(x2+y2)2=x2−y2
(x2+y2)2.
Step 4: Evaluate the double integral.
ZZDy2−x2
(x2+y2)2−x2−y2
(x2+y2)2dA =ZZD
0dA = 0.
Therefore, the value of the line integral HC
y
x2+y2dx+x
x2+y2dy using Green’s
Theorem is 0.
Question 19
Question
Let Cbe the circle centered at the origin with radius 2 oriented counterclockwise,
and let Dbe the region enclosed by C. Given the vector field F(x, y)=(x2+
y2,−x), calculate the line integral of Faround the boundary of Dusing Green’s
Theorem.
18
Solution
Step 1: Compute ∂Q
∂x and ∂P
∂y .
∂Q
∂x =∂
∂x(−x) = −1
∂P
∂y =∂
∂y (x2+y2) = 2y
Step 2: Apply Green’s Theorem, which states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
Step 3: Evaluate the double integral over D.
ZZD
(−1−2y)dA =−ZZD
(1 + 2y)dA
Step 4: Convert to polar coordinates since Cis a circle. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =rdrdθ, with rranging from 0 to 2 and θ
ranging from 0 to 2π.
Step 5: Replace xand yin the integral with rcos(θ)and rsin(θ).
−Z2π
0Z2
0
(1 + 2rsin(θ))rdrdθ
Step 6: Evaluate the integral.
−Z2π
0Z2
0
r+ 2r2sin(θ)drdθ
−Z2π
0 r2
2+2r3
3sin(θ)
2
0!dθ
−Z2π
02 + 16
3sin(θ)dθ
−2θ−16
3cos(θ)
2π
0
−(4π−0) = −4π
Therefore, the line integral of Faround the boundary of Dis −4π.
Question 20
Question
Let Cbe the curve given by the intersection of the cylinder x2+y2= 1 and
the plane 2x+y+z= 3. Use Green’s Theorem to evaluate the line integral
HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).
19
Solution
To apply Green’s Theorem, we first need to find a region Din the xy-plane
that contains the curve Cas its boundary. Since Cis the intersection of the
cylinder x2+y2= 1 and the plane 2x+y+z= 3, we can rewrite zin terms
of xand yas z= 3 −2x−y. So the projection of Conto the xy-plane is the
circle x2+y2= 1.
Let Dbe the disk enclosed by the circle x2+y2= 1. Then we have ∂D =C.
By Green’s Theorem, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z)) = (y+z, x +z, x +y).
Now, we compute:
∂Q
∂x = 1,
∂P
∂y = 1,
∂Q
∂x −∂P
∂y = 0.
Therefore, the line integral can be simplified as:
IC
F·dr=ZZD
0dA = 0.
So, the value of the line integral is 0.
Question 21
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Let
F(x, y)=(x2+y2)i+x2j. Calculate the line integral HCF·dr using Green’s
Theorem.
Solution
Step 1: Find the region Denclosed by the curve C. Since x2+y2= 4 represents
a circle of radius 2 centered at the origin, the region Denclosed by Cis also
the region inside this circle.
Step 2: Calculate ∂Q
∂x and ∂P
∂y . Given F(x, y)=(x2+y2)i+x2j, we have
P(x, y) = x2+y2and Q(x, y) = x2.
Calculating the partial derivatives:
∂Q
∂x = 2xand ∂P
∂y = 2y
20
Step 3: Apply Green’s Theorem. Green’s Theorem states: HCF·dr =
RRD∂Q
∂x −∂P
∂y dA
Substitute the partial derivatives:
IC
F·dr =ZZD
(2x−2y)dA
Step 4: Convert the integral to polar coordinates. In polar coordinates,
dA =rdrdθ and x=rcos(θ), y =rsin(θ).
The integral becomes:
IC
F·dr =Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ
Step 5: Evaluate the integral.
IC
F·dr =Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ
=Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=−16
3sin(θ)−16
3cos(θ)2π
0
=−16
3(sin(2π) + cos(2π)−sin(0) −cos(0))
=−16
3(0 + 1 −0−1) = 32
3
Therefore, the line integral HCF·dr =32
3.
Question 22
Question
Let Cbe the curve defined by x(t) = 3 cos(t),y(t) = 2 sin(t)for 0≤t≤2π.
Evaluate the line integral HC(x2+y2)dx +xy dy.
21
Solution
Step 1: Calculate the derivatives of x(t)and y(t)with respect to t.
dx
dt =−3 sin(t)and dy
dt = 2 cos(t)
Step 2: Compute dx and dy.
dx =dx
dt dt =−3 sin(t)dt and dy =dy
dt dt = 2 cos(t)dt
Step 3: Substitute x(t),y(t),dx, and dy into the line integral.
IC
(x2+y2)dx+xy dy =Z2π
0
((3 cos(t))2+(2 sin(t))2)(−3 sin(t))+3 cos(t)·2 cos(t)dt
Step 4: Simplify the integrand.
Z2π
0
(9 cos2(t)+4 sin2(t))(−3 sin(t))+6 cos(t) sin(t)dt =Z2π
0
(−27 cos2(t) sin(t)−12 sin3(t)+6 cos(t) sin(t)) dt
Step 5: Evaluate the integral.
=−27 Z2π
0
cos2(t) sin(t)dt −12 Z2π
0
sin3(t)dt + 6 Z2π
0
cos(t) sin(t)dt
Step 6: Use trigonometric identities to simplify and evaluate each integral.
=−27 −cos3(t)
32π
0
−12 −cos2(t)
2+cos4(t)
42π
0
+ 6 −cos2(t)
22π
0
Step 7: Substitute 2πand 0into the expressions and simplify to find the
final answer.
Question 23
Question
Let Cbe the positively oriented circle given by x2+y2= 4. Evaluate the line
integral HC(y3dx +x3dy)using Green’s Thejsonreorem.
Solution
Step 1: We need to express the line integral as a double integral over the region
enclosed by Cusing Green’s Theorem. Green’s Theorem relates the line integral
of a vector field over a closed curve to the double integral of the divergence of a
related vector field over the region enclosed by the curve. The theorem states:
IC
(P dx +Qdy) = ZZR∂Q
∂x −∂P
∂y dA
22
Step 2: The given line integral HC(y3dx +x3dy)can be written as HCx3dy +
y3dx. Let P(x, y) = x3and Q(x, y) = y3. Then, ∂Q
∂x = 0 and ∂P
∂y = 0. Hence,
the double integral reduces to zero.
Step 3: Since the double integral over the region enclosed by Cis zero, the
value of the line integral HC(y3dx +x3dy)is also zero.
Therefore, the value of the line integral HC(y3dx +x3dy)over the circle
x2+y2= 4 is 0.
Question 24
Question
Evaluate the line integral HC(x2+y2)dx+2xydy, where Cis the circle x2+y2= 4,
oriented counterclockwise.
Solution
Step 1: We will first rewrite the given line integral in terms of a potential
function. Let P(x, y) = x2+y2and Q(x, y) = 2xy. Then, we have ∂Q
∂x = 2y=
∂P
∂y , which implies that the integrand is exact. By Green’s Theorem, we have
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by C, which is the circle x2+y2= 4.
Step 2: The region Dcan be described using the polar coordinates x=
rcos(θ),y=rsin(θ), with 0≤r≤2and 0≤θ≤2π. Thus, dA =rdrdθ. The
integral becomes
ZZD∂Q
∂x −∂P
∂y dA =Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ.
Step 3: We can now evaluate the double integral:
Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ =Z2π
0Z2
0
2r2sin(θ)−2r2drdθ
=Z2π
0 2
3r3sin(θ)−r3r=2
r=0!dθ
=Z2π
016
3sin(θ)−8dθ
=−16
3cos(θ)−8θ2π
0
=−16
3(cos(2π)−cos(0)) −8(2π−0)
=−16
3−16π.
23
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
circle x2+y2= 4 oriented counterclockwise is −16
3−16π.
Question 25
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral
IC
(x2−y2)dx + 2xy dy.
Solution
Step 1: First, let’s calculate the partial derivatives of the given vector field
F(x, y) = (x2−y2,2xy).
∂
∂y (x2−y2) = −2y, ∂
∂x(2xy) = 2y.
Step 2: Now, Green’s Theorem states that for a simply connected region
Dbounded by a positively oriented, piecewise-smooth, simple closed curve C,
the line integral of a vector field
F= (P, Q)along Ccan be evaluated as
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where Pand Qare the component functions of
F.
Step 3: Applying Green’s Theorem to the given line integral, we have
IC
(x2−y2)dx + 2xy dy =ZZD
(2y−(−2y)) dA = 4 ZZD
y dA.
Step 4: The region Denclosed by the curve Cis a circle of radius 2.
We can use polar coordinates to evaluate the double integral. The conversion
formula from Cartesian to polar coordinates is x=rcos θand y=rsin θ, with
dA =r dr dθ.
Step 5: Substituting y=rsin θand dA =r dr dθ into the double integral,
we get
4ZZD
y dA = 4 Z2π
0Z2
0
rsin θ·r dr dθ.
24
Step 6: Evaluating the double integral, we have
4Z2π
0Z2
0
rsin θ·r dr dθ = 4 Z2π
0r3cos θ
3
2
0sin θ dθ
= 4 Z2π
0
8
3cos θsin θ dθ
=32
3Z2π
0
1
2sin(2θ)dθ
=32
3−1
4cos(2θ)
2π
0
=32
3−1
4−(−1
4)
=32
3×0
= 0.
Therefore, the value of the line integral HC(x2−y2)dx + 2xy dy along the
curve Cis 0.
25
Step 5: Convert the double integral to polar coordinates since Dis a disk:
ZZDx2−2x2y+ydA =Z2π
0Z1
0(rcos θ)2−2(rcos θ)2(rsin θ)+(rsin θ)r drdθ
Step 6: Evaluate the double integral:
=Z2π
0Z1
0r2cos2θ−2r3cos2θsin θ+rsin θr drdθ
=Z2π
0Z1
0r3(cos2θ−2rcos2θsin θ+ sin θ)drdθ
=Z2π
0r4
4(cos2θ−2rcos2θsin θ+ sin θ)1
0
dθ
=Z2π
01
4(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
4Z2π
0
(cos2θ−2 cos2θsin θ+ sin θ)dθ
=1
41
2(θ+ sin θcos θ−2 sin2θ)2π
0
=1
41
2(2π+ 0 −0)
Question 2
Question
Let Cbe the curve defined by x=t3and y=t2for 0≤t≤1. Use Green’s
Theorem to evaluate HC(x2+y2)dx +xy dy.
Solution
Step 1: We start by parameterizing the curve Cusing x=t3and y=t2. Then
dx = 3t2dt and dy = 2t dt. The curve Cbecomes {(t3, t2) : 0 ≤t≤1}.
Step 2: We rewrite the line integral HC(x2+y2)dx +xy dy using the param-
eterization:
IC
(x2+y2)dx +xy dy =Z1
0
((t3)2+ (t2)2)·3t2dt + (t3·t2)·2t dt
Step 3: Simplify the integrand:
Z1
0
(t6+t4)·3t2dt + 2t4dt =Z1
0
(3t8+ 3t6)dt + 2t4dt
2
Step 4: Integrate the terms one by one:
Z1
0
3t8dt +Z1
0
3t6dt +Z1
0
2t4dt
Step 5: Calculate the integrals:
3
9t91
0
+3
7t71
0
+2
5t51
0
Step 6: Substitute the upper and lower limits and simplify the expression:
3
9+3
7+2
5=1
3+3
7+2
5=35 + 45 + 42
105 =122
105
Step 7: Therefore, the value of the line integral is 122
105 .
Question 3
Question
Let Cbe the curve formed by the intersection of the plane z=x+yand
the cylinder x2+y2= 1, oriented counterclockwise when viewed from above.
Use Green’s Theorem to evaluate the line integral HCF·drwhere F(x, y, z) =
⟨x2, y2, z2⟩.
Solution
To evaluate the line integral using Green’s Theorem, we first need to find the
curve Cthat bounds the region D.
Step 1: Find the Region D
The curve Cis the intersection of the plane z=x+yand the cylinder x2+y2= 1.
Solving for zin terms of xand y, we have z=x+y. This gives us the region
Dbounded by the circle x2+y2= 1 in the xy-plane.
Step 2: Calculate the Line Integral Using Green’s Theorem
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
where F(x, y, z) = ⟨P, Q, R⟩. In this case, F(x, y, z) = ⟨x2, y2, z2⟩, so P=x2,
Q=y2, and R=z2.
The partial derivatives are: ∂Q
∂x = 0
∂P
∂y = 0
3
Thus, the line integral simplifies to:
IC
F·dr=ZZD
(0 −0)dA = 0
Therefore, the value of the line integral HCF·dris 0.
Question 4
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using Green’s Theorem.
Solution
Step 1: Determine the region Denclosed by the curve C. The curve Cis a
circle centered at the origin with radius 2. Therefore, the region Denclosed by
Cis the interior of the circle.
Step 2: Check if Green’s Theorem applies. Green’s Theorem states that for
a simple, closed, positively oriented curve Cenclosing a region D, if P(x, y)
and Q(x, y)have continuous partial derivatives on an open region containing D,
then IC
P dx +Qdy =ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Verify the conditions for Green’s Theorem. Given P(x, y) = 3x2y−y3
and Q(x, y) = x3−2xy2,
∂Q
∂x = 3x2−2y2and ∂P
∂y = 3x2−3y2.
Both ∂Q
∂x and ∂P
∂y are continuous on D, so Green’s Theorem applies.
Step 4: Apply Green’s Theorem. The line integral can be written as
IC
(3x2y−y3)dx + (x3−2xy2)dy =ZZD∂(x3−2xy2)
∂x −∂(3x2y−y3)
∂y dA.
Simplifying the right-hand side gives
ZZD
(3x2−2y2−3x2+ 3y2)dA =ZZD
y2dA.
Step 5: Calculate the double integral. Since the region Dis a circle of radius
2 centered at the origin, we can convert the integral to polar coordinates with
0≤r≤2and 0≤θ≤2π. Therefore, the double integral becomes
Z2π
0Z2
0
r2·rdrdθ.
4
Step 6: Evaluate the integral. Computing the double integral gives
Z2π
0Z2
0
r3drdθ =Z2π
0r4
42
0
dθ =Z2π
0
4dθ = 8π.
Hence, the value of the line integral HC(3x2y−y3)dx + (x3−2xy2)dy using
Green’s Theorem is 8π.
Question 5
Question
Let Cbe the curve formed by the intersection of the plane 2x+y+z= 4 and
the cylinder x2+y2= 1. Use Green’s Theorem to evaluate the line integral
HC(y2+z2)dx + (z2+x2)dy + (x2+y2)dz.
Solution
To use Green’s Theorem, we first need to find a vector field Fsuch that ∇×F=
y2+z2i+z2+x2j+x2+y2k.
Let’s parameterize the curve C. We can use cylindrical coordinates to de-
scribe Cas x= cos(t), y = sin(t), z = 4 −2 cos(t)−sin(t)where 0≤t≤2π.
Step 1: Calculate the partial derivatives of x,y, and zwith respect to t.
dx
dt =−sin(t),dy
dt = cos(t),dz
dt = 2 sin(t)−cos(t)
Step 2: Calculate the line integral. Using Green’s Theorem, we have
IC
F·dr=ZZD
(∇ × F)·ndA
Where Dis the region enclosed by curve Cand nis the outward unit normal
to D.
Now we need to find the unit normal vector n=1
√6(−1,−2,1).
Step 3: Calculate the curl of F.
∇ × F=
i j k
∂
∂x
∂
∂y
∂
∂z
y2+z2z2+x2x2+y2
= (2y−2y, 2z−2z, 2x−2x) = 0
Since the curl is zero, the line integral over Cis 0.
Question 6
Question
Let Cbe the curve consisting of the line segment from (0,0) to (2,2) followed by
the line segment from (2,2) to (4,0) traversed in the counterclockwise direction.
5
Let Dbe the region enclosed by C. Use Green’s Theorem to evaluate the line
integral HCF·dr, where F(x, y) = (y2,sin(xy)).
Solution
Using Green’s Theorem, we have that
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = (P, Q).
Step 1: Compute the partial derivatives of Pand Q
∂Q
∂x =∂
∂x(sin(xy)) = ycos(xy),
∂P
∂y =∂
∂y (y2) = 2y.
Step 2: Evaluate the double integral over region DUsing the given curve C,
we can parameterize the line segments as follows: For the first segment: x=t,
y=t,0≤t≤2. For the second segment: x= 2t+ 2,y=−2t+ 2,0≤t≤1.
Therefore, the integral becomes:
ZZD
(ycos(xy)−2y)dA
=Z2
0Zt
0
(tcost2−2t)dy dx +Z1
0Z−2t+2
2t+2
((−2t+ 2) cos(−2t(2t+ 2)) −2(−2t+ 2)) dy dx
=Z2
0 t2sint2
2−t2!dx +Z1
0
(2t(2 cos(2t(2t+ 2)) −2) −2(2t−2)) dx.
This integral may be simplified further, but the final answer depends on the
specific values obtained after solving the nested integrals.
Question 7
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(2x−y)dx + (x+y)dy using Green’s Theorem.
Solution
1. We first need to determine the region enclosed by the curve C, which is a
circle centered at the origin with radius 2. Let Dbe the region enclosed
by C.
6
2. Green’s Theorem states that for a simply connected region Dbounded by a
positively-oriented, piecewise-smooth, simple closed curve C, if
F(x, y) =
(P(x, y), Q(x, y)) is a continuously differentiable vector field defined on an
open region containing D, then
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where dr =dxi+dyjand dA =dxdy.
3. In our case,
F(x, y) = (2x−y, x+y),P(x, y) = 2x−y, and Q(x, y) = x+y.
We can rewrite the line integral as
IC
(2x−y)dx + (x+y)dy =ZZD
(2 −1) dA =ZZD
dA.
4. To evaluate RRDdA, we convert the double integral into polar coordinates.
Let x=rcos θand y=rsin θ. Then dA =r dr dθ, and the integral
becomes
ZZD
dA =Z2π
0Z2
0
r dr dθ.
5. Solving the inner integral, we have
Z2
0
r dr =r2
2
2
0= 4.
6. Substituting this back into the double integral, we get
Z2π
0Z2
0
r dr dθ =Z2π
0
4dθ = 8π.
7. Therefore, the line integral HC(2x−y)dx + (x+y)dy is equal to 8π.
Question 8
Question
Let Cbe the curve given by the intersection of the surfaces z=x2and y=x3,
oriented counterclockwise when viewed from above. Use Green’s Theorem to
evaluate the line integral HC(x2+y2)dx + 2xydy.
Solution
Step 1: Determine the region Denclosed by the curve C. To find the region
Denclosed by C, we need to find the intersection points of the surfaces z=x2
and y=x3. Solving the equations z=x2and y=x3simultaneously, we get
7
z=x2and z=y2/3. So Dlies between the curves y=x3and y=x2/3. Now,
we can express the boundaries of Das x=y1/3and x=y2/3.
Step 2: Calculate the partial derivatives. Let P(x, y) = x2+y2and
Q(x, y) = 2xy. We need to compute ∂Q
∂x −∂P
∂y .
∂Q
∂x = 2yand ∂P
∂y = 2y
Step 3: Apply Green’s Theorem. By Green’s Theorem,
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA
=ZZD
(2y−2y)dA
=ZZD
0dA
= 0
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
curve Cis 0.
Question 9
Question
Let Cbe the curve defined by C(t) = (sin t, cos t)for 0≤t≤π/2. Calculate
the line integral HCF·dr, where F(x, y) = (−y, x), using Green’s Theorem.
Solution
Step 1: Find the region Denclosed by the curve C.
• We have C(t) = (sin t, cos t)for 0≤t≤π/2which represents the quarter-
circle of radius 1 centered at the origin in the first quadrant.
• Therefore, the region Denclosed by Cis the area of this quarter-circle.
Step 2: Verify if the hypotheses of Green’s Theorem are satisfied.
• Green’s Theorem states that for a positively oriented simple closed curve
C, the line integral of Falong Cis equal to the double integral of (∂Q/∂x−
∂P/∂y)over the region Denclosed by C, where F= (P, Q).
• Calculate ∂Q
∂x −∂P
∂y for F(x, y) = (−y, x):
∂Q
∂x −∂P
∂y = 1 −(−1) = 2
This is a nonzero constant, which is good.
8
Step 3: Apply Green’s Theorem to compute the line integral using the double
integral.
• Green’s Theorem gives
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
=ZZD
2dA
= 2 ZZD
dA
= 2 ·(Area of D)
Since the area of the quarter-circle is 1
4π(1)2=π
4, the line integral is
2·π
4=π
2.
Thus, the line integral HCF·dr=π
2.
Question 10
Question
Let Cbe the closed curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral HC(y2−x)dx +x2dy.
Solution
Step 1: Let’s first find the region Denclosed by the curve C, which is the circle
with radius 2 centered at the origin. We can see that Dis the disk x2+y2≤4.
Step 2: Green’s Theorem states that for a positively oriented simple closed
curve Cand a region Dwhose boundary is C, if P(x, y)and Q(x, y)have
continuous partial derivatives on an open region containing D, then
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA.
Step 3: Let P(x, y) = y2−xand Q(x, y) = x2. Then, we have
∂Q
∂x = 2xand ∂P
∂y = 2y.
Step 4: We can rewrite the line integral as
IC
(y2−x)dx +x2dy =ZZD
(2x−2y)dA.
9
Step 5: Since Dis the disk of radius 2 centered at the origin, we can switch
to polar coordinates by letting x=rcos(θ)and y=rsin(θ). Then dA =rdrdθ,
and the integral becomes
Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ.
Step 6: This simplifies to
Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ =Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=16
3[sin(θ) + cos(θ)]2π
0
=16
3[sin(2π) + cos(2π)−(sin(0) + cos(0))]
=16
3(0 + 1 −(0 + 1))
= 0 .
Therefore, the value of the line integral HC(y2−x)dx +x2dy is 0.
Question 11
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise, and let D
be the region enclosed by C. Use Green’s Theorem to evaluate the line integral
HC(x2+y2)dx +xydy.
Solution
Step 1: First, let’s calculate the partial derivatives of N(x, y) = x2+y2and
M(x, y) = xy.
Given M(x, y) = xy, we have:
∂M
∂y =x
Given N(x, y) = x2+y2, we have:
∂N
∂x = 2x
Step 2: Now, apply Green’s Theorem, which states:
IC
(Mdx +N dy) = ZZD∂N
∂x −∂M
∂y dA
10
So, we have:
IC
(x2+y2)dx +xydy =ZZD
(2x−x)dA
Step 3: Simplify the double integral:
ZZD
(2x−x)dA =ZZD
xdA
Step 4: The region Denclosed by the curve Cis a circle with radius 2. Thus,
we can convert the double integral to polar coordinates:
ZZD
xdA =Z2π
0Z2
0
(rcos θ)rdrdθ
Step 5: Evaluate the integral in polar coordinates:
Z2π
0Z2
0
(rcos θ)rdrdθ =Z2π
023
3cos θdθ =16
3π
Therefore, the value of the line integral HC(x2+y2)dx +xydy is 16
3π.
Question 12
Question
Let Cbe the curve given by the intersection of the plane z= 4 −xand the
cylinder x2+y2= 4. Use Green’s Theorem to find the circulation of the vector
field F= (y2, z, x)around the curve C.
Solution
To apply Green’s Theorem, we need to find the normal vector nto the curve C.
Step 1: Find the parametric equations for the curve C.
The intersection of the plane z= 4 −xand the cylinder x2+y2= 4 can be
parameterized as follows:
x= 2 cos(t), y = 2 sin(t), z = 4 −2 cos(t)
for 0≤t≤2π.
Step 2: Calculate the circulation of the vector field F.
The circulation of Faround Cis given by the line integral:
IC
F·dr=ZC
F·Tds
where Tis the unit tangent vector to the curve C.
11
The unit tangent vector Tis given by dr
dt . Hence, we have:
T= (−2 sin(t),2 cos(t),2 sin(t))
Now, we can compute the line integral:
IC
F·dr=Z2π
0
F(2 sin(t),4−2 cos(t),2 cos(t)) ·(−2 sin(t),2 cos(t),2 sin(t)) dt
Calculating this line integral will give us the circulation of the vector field F
around the curve C.
Question 13
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Calculate
the line integral HC(x2y−xy2)dx + (x3y−y3)dy using Green’s Theorem.
Solution
Step 1: Determine the region enclosed by the curve C. We can see that Cis a
circle of radius 2centered at the origin.
Step 2: Write the given line integral in terms of a double integral by apply-
ing Green’s Theorem. Green’s Theorem states that for a vector field F= (P, Q),
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by the curve C.
In this case, F= (x2y−xy2, x3y−y3), and we have:
P=x2y−xy2,
Q=x3y−y3.
Calculating the partial derivatives:
∂Q
∂x = 3x2y,
∂P
∂y =x2−2xy.
So, the line integral becomes:
IC
(x2y−xy2)dx + (x3y−y3)dy =ZZD
(3x2y−x2+ 2xy)dA.
12
Step 3: Find the integral over the region D. Since Dis a disk of radius 2,
we can express Din polar coordinates as 0≤r≤2,0≤θ≤2π. Therefore, we
have:
ZZD
(3x2y−x2+ 2xy)dA =Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ.
Step 4: Evaluate the double integral. Let’s simplify the integrand:
3r3sin2(θ)−r2+ 2rsin(θ) = 3r4sin2(θ)−r2+ 2rsin(θ).
Then, compute the double integral:
Z2π
0Z2
0
(3r3sin2(θ)−r2+ 2rsin(θ))r drdθ
=Z2π
03
5r5sin2(θ)−1
3r3+r2sin(θ)
2
0
dθ
=Z2π
096
5sin2(θ)−8
3+ 4 sin(θ)dθ
=96
5θ
2−sin(2θ)
4−8
3θ−4 cos(θ)
2π
0
= 0.
Step 5: Conclusion. Therefore, the line integral HC(x2y−xy2)dx + (x3y−
y3)dy around the curve Cis equal to 0.
Question 14
Question
Let Cbe the curve consisting of the line segment from (1,0) to (0,1), followed
by the arc of the circle x2+y2= 1 from (0,1) to (1,0). Use Green’s Theorem
to evaluate the line integral HC(x2−y)dx +xydy.
Solution
Step 1: First, we need to parameterize the curve C. Let’s denote the line
segment as C1and the arc of the circle as C2. We can parameterize C1as
r1(t) = (1 −t, t)for 0≤t≤1, and C2as r2(t) = (cos(t),sin(t)) for 0≤t≤π.
Step 2: Next, we need to compute the partial derivatives of P(x, y) = x2−y
and Q(x, y) = xy. We have:
∂Q
∂x =yand ∂P
∂y =−1
13
Step 3: Now, we can apply Green’s Theorem, which states that for a counterclockwise-
oriented simple closed curve Cand a region Dbounded by C, we have:
IC
(P dx +Qdy) = ZZD∂Q
∂x −∂P
∂y dA
Step 4: Substituting the given Pand Q, and the computed partial deriva-
tives, into Green’s Theorem, we get:
IC
(x2−y)dx +xydy =ZZD
(y+ 1)dA
Step 5: Since Dis the unit disk, we can rewrite the double integral in polar
coordinates. Thus, the line integral becomes:
IC
(x2−y)dx +xydy =Z2π
0Z1
0
(rsin(θ) + 1)rdrdθ
Step 6: Solving the double integral, we obtain:
=Z2π
0Z1
0
r2sin(θ) + rdrdθ
=Z2π
01
2sin(θ) + 1
2dθ
=−1
2cos(θ) + θ
22π
0
=π
Thus, the value of the line integral HC(x2−y)dx +xydy around the curve C
is π.
Question 15
Question
Let Cbe the square with vertices at (0,0),(2,0),(2,2), and (0,2). Evaluate
the line integral HC(x2y−y2)dx + (x2+xy)dy using Green’s theorem.
Solution
Step 1: Calculate the partial derivatives of the given vector field F(x, y) =
(x2y−y2, x2+xy).
∂
∂y (x2y−y2) = x2−2y
∂
∂x(x2+xy) = 2x+y
14
Step 2: Compute the double integrals over the region Denclosed by the
square C.
ZZD∂Q
∂x −∂P
∂y dA
ZZD(2x+y)−(x2−2y)dA
Step 3: Find the limits of integration for the double integral. The square C
can be parameterized by x=t, y =tfor 0≤t≤2.
Step 4: Evaluate the double integral.
Z2
0Z2
0
((2x+y)−(x2−2y))dydx
Z2
0Z2
0
(2t+t−t2+ 2t)dydx
Z2
0Z2
0
(5t−t2)dydx
Z2
0
[(5t−t2)y]2
0dx
Z2
0
(10 −4t)dx
Z2
0
10dx −Z2
0
4tdx
10(2) −422
2−02
2
20 −4(2)
20 −8 = 12
Therefore, the value of the line integral HC(x2y−y2)dx + (x2+xy)dy using
Green’s theorem is 12 .
Question 16
Question
Let Cbe the circle given by the curve parameterization r(t) = (3 cos t, 3 sin t)
where 0≤t≤2π. Calculate the circulation HC
f·dr for the vector field
f(x, y) = (−y, x)using Green’s Theorem.
15
Solution
Step 1: Calculate the circulation directly:
IC
f·dr =Z2π
0
f(r(t)) ·r′(t)dt
=Z2π
0
(−3 sin t, 3 cos t)·(−3 sin t, 3 cos t)dt
=Z2π
0
9 sin2t+ 9 cos2tdt
=Z2π
0
9dt
= 18π
Step 2: Use Green’s Theorem to calculate the circulation: Green’s Theorem
states that for a region Renclosed by a simple closed curve C, and a vector
field
F=Pi+Qjdefined on an open region containing R:
IC
F·dr =ZZR∂Q
∂x −∂P
∂y dA
Step 3: Validate Green’s Theorem for our vector field
f(x, y) = (−y, x): Let
P(x, y) = −yand Q(x, y) = x, then:
∂Q
∂x = 1,∂P
∂y =−1
∂Q
∂x −∂P
∂y = 2
Step 4: Green’s Theorem gives:
IC
f·dr =ZZR
2dA
= 2 ·Area(R)
Step 5: The circle Cencloses the region Rcorresponding to the disk x2+y2≤
9. The area of this disk is π·32= 9π. Therefore:
IC
f·dr = 2 ·9π= 18π
Thus, the circulation of the vector field
f(x, y)=(−y, x)around the circle
Cis 18π.
16
Question 17
Question
Let Cbe the curve given by x2+y2= 4 oriented counterclockwise, and let D
be the region bounded by C. Use Green’s Theorem to evaluate the line integral
IC
(x2y−y2)dx + (x2+xy)dy.
Solution
Step 1: Identify the region Dbounded by the curve C.The curve Cis
a circle centered at the origin with radius 2. Therefore, the region Dbounded
by Cis the interior of the circle.
Step 2: Apply Green’s Theorem. Green’s Theorem states that for
a region Dbounded by a positively oriented, piecewise-smooth, simple closed
curve C, and for a vector field F(x, y)=(P(x, y), Q(x, y)) with continuous
partial derivatives on an open region containing D, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where dr= (dx, dy)is the differential path element and dA =dx dy is the area
element.
In our case, P(x, y) = x2y−y2and Q(x, y) = x2+xy. We want to calculate
the line integral HCF·dr, which is equal to the double integral RRD∂Q
∂x −∂P
∂y dA.
Step 3: Calculate the double integral. We first need to find ∂Q
∂x and
∂P
∂y :
∂Q
∂x = 2x+yand ∂P
∂y =x2−2y.
Therefore, the expression ∂Q
∂x −∂P
∂y = 2x+y−x2+ 2y.
Step 4: Evaluate the double integral over D.Now, we can proceed to
evaluate the double integral over the region D:
ZZD
(2x+y−x2+ 2y)dA =ZZD
(2x−x2+ 3y)dA.
To simplify this integral, we can convert to polar coordinates (x=rcos θ,
y=rsin θ) with dA =r dr dθ:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ.
Step 5: Compute the integral. Evaluating the integral, we get:
Z2π
0Z2
0
(2rcos θ−r2cos2θ+ 3rsin θ)r dr dθ =π.
So, the value of the line integral HC(x2y−y2)dx+(x2+xy)dy over the curve
Cis π.
17
Question 18
Question
Let Cbe the curve defined by x2+y2= 4, oriented counterclockwise. Using
Green’s Theorem, evaluate the line integral HC
y
x2+y2dx +x
x2+y2dy.
Solution
Step 1: Determine the region enclosed by the curve C. The curve Cis a circle
with radius 2 centered at the origin. The region enclosed by Cis the interior of
this circle.
Step 2: Express the line integral as a double integral using Green’s Theorem.
Green’s Theorem states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y) = y
x2+y2,x
x2+y2and Dis the region enclosed by C.
In this case, P(x, y) = y
x2+y2and Q(x, y) = x
x2+y2.
Step 3: Compute the partial derivatives.
∂Q
∂x =(x2+y2)−x(2x)
(x2+y2)2=y2−x2
(x2+y2)2,
∂P
∂y =(x2+y2)−y(2y)
(x2+y2)2=x2−y2
(x2+y2)2.
Step 4: Evaluate the double integral.
ZZDy2−x2
(x2+y2)2−x2−y2
(x2+y2)2dA =ZZD
0dA = 0.
Therefore, the value of the line integral HC
y
x2+y2dx+x
x2+y2dy using Green’s
Theorem is 0.
Question 19
Question
Let Cbe the circle centered at the origin with radius 2 oriented counterclockwise,
and let Dbe the region enclosed by C. Given the vector field F(x, y)=(x2+
y2,−x), calculate the line integral of Faround the boundary of Dusing Green’s
Theorem.
18
Solution
Step 1: Compute ∂Q
∂x and ∂P
∂y .
∂Q
∂x =∂
∂x(−x) = −1
∂P
∂y =∂
∂y (x2+y2) = 2y
Step 2: Apply Green’s Theorem, which states:
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA
Step 3: Evaluate the double integral over D.
ZZD
(−1−2y)dA =−ZZD
(1 + 2y)dA
Step 4: Convert to polar coordinates since Cis a circle. In polar coordinates,
x=rcos(θ),y=rsin(θ), and dA =rdrdθ, with rranging from 0 to 2 and θ
ranging from 0 to 2π.
Step 5: Replace xand yin the integral with rcos(θ)and rsin(θ).
−Z2π
0Z2
0
(1 + 2rsin(θ))rdrdθ
Step 6: Evaluate the integral.
−Z2π
0Z2
0
r+ 2r2sin(θ)drdθ
−Z2π
0 r2
2+2r3
3sin(θ)
2
0!dθ
−Z2π
02 + 16
3sin(θ)dθ
−2θ−16
3cos(θ)
2π
0
−(4π−0) = −4π
Therefore, the line integral of Faround the boundary of Dis −4π.
Question 20
Question
Let Cbe the curve given by the intersection of the cylinder x2+y2= 1 and
the plane 2x+y+z= 3. Use Green’s Theorem to evaluate the line integral
HCF·dr, where F(x, y, z) = (y+z, x +z, x +y).
19
Solution
To apply Green’s Theorem, we first need to find a region Din the xy-plane
that contains the curve Cas its boundary. Since Cis the intersection of the
cylinder x2+y2= 1 and the plane 2x+y+z= 3, we can rewrite zin terms
of xand yas z= 3 −2x−y. So the projection of Conto the xy-plane is the
circle x2+y2= 1.
Let Dbe the disk enclosed by the circle x2+y2= 1. Then we have ∂D =C.
By Green’s Theorem, we have
IC
F·dr=ZZD∂Q
∂x −∂P
∂y dA,
where F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z)) = (y+z, x +z, x +y).
Now, we compute:
∂Q
∂x = 1,
∂P
∂y = 1,
∂Q
∂x −∂P
∂y = 0.
Therefore, the line integral can be simplified as:
IC
F·dr=ZZD
0dA = 0.
So, the value of the line integral is 0.
Question 21
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Let
F(x, y)=(x2+y2)i+x2j. Calculate the line integral HCF·dr using Green’s
Theorem.
Solution
Step 1: Find the region Denclosed by the curve C. Since x2+y2= 4 represents
a circle of radius 2 centered at the origin, the region Denclosed by Cis also
the region inside this circle.
Step 2: Calculate ∂Q
∂x and ∂P
∂y . Given F(x, y)=(x2+y2)i+x2j, we have
P(x, y) = x2+y2and Q(x, y) = x2.
Calculating the partial derivatives:
∂Q
∂x = 2xand ∂P
∂y = 2y
20
Step 3: Apply Green’s Theorem. Green’s Theorem states: HCF·dr =
RRD∂Q
∂x −∂P
∂y dA
Substitute the partial derivatives:
IC
F·dr =ZZD
(2x−2y)dA
Step 4: Convert the integral to polar coordinates. In polar coordinates,
dA =rdrdθ and x=rcos(θ), y =rsin(θ).
The integral becomes:
IC
F·dr =Z2π
0Z2
0
(2rcos(θ)−2rsin(θ))rdrdθ
Step 5: Evaluate the integral.
IC
F·dr =Z2π
0Z2
0
(2r2cos(θ)−2r2sin(θ))drdθ
=Z2π
02
3r3cos(θ)−2
3r3sin(θ)2
0
dθ
=Z2π
016
3cos(θ)−16
3sin(θ)dθ
=−16
3sin(θ)−16
3cos(θ)2π
0
=−16
3(sin(2π) + cos(2π)−sin(0) −cos(0))
=−16
3(0 + 1 −0−1) = 32
3
Therefore, the line integral HCF·dr =32
3.
Question 22
Question
Let Cbe the curve defined by x(t) = 3 cos(t),y(t) = 2 sin(t)for 0≤t≤2π.
Evaluate the line integral HC(x2+y2)dx +xy dy.
21
Solution
Step 1: Calculate the derivatives of x(t)and y(t)with respect to t.
dx
dt =−3 sin(t)and dy
dt = 2 cos(t)
Step 2: Compute dx and dy.
dx =dx
dt dt =−3 sin(t)dt and dy =dy
dt dt = 2 cos(t)dt
Step 3: Substitute x(t),y(t),dx, and dy into the line integral.
IC
(x2+y2)dx+xy dy =Z2π
0
((3 cos(t))2+(2 sin(t))2)(−3 sin(t))+3 cos(t)·2 cos(t)dt
Step 4: Simplify the integrand.
Z2π
0
(9 cos2(t)+4 sin2(t))(−3 sin(t))+6 cos(t) sin(t)dt =Z2π
0
(−27 cos2(t) sin(t)−12 sin3(t)+6 cos(t) sin(t)) dt
Step 5: Evaluate the integral.
=−27 Z2π
0
cos2(t) sin(t)dt −12 Z2π
0
sin3(t)dt + 6 Z2π
0
cos(t) sin(t)dt
Step 6: Use trigonometric identities to simplify and evaluate each integral.
=−27 −cos3(t)
32π
0
−12 −cos2(t)
2+cos4(t)
42π
0
+ 6 −cos2(t)
22π
0
Step 7: Substitute 2πand 0into the expressions and simplify to find the
final answer.
Question 23
Question
Let Cbe the positively oriented circle given by x2+y2= 4. Evaluate the line
integral HC(y3dx +x3dy)using Green’s Thejsonreorem.
Solution
Step 1: We need to express the line integral as a double integral over the region
enclosed by Cusing Green’s Theorem. Green’s Theorem relates the line integral
of a vector field over a closed curve to the double integral of the divergence of a
related vector field over the region enclosed by the curve. The theorem states:
IC
(P dx +Qdy) = ZZR∂Q
∂x −∂P
∂y dA
22
Step 2: The given line integral HC(y3dx +x3dy)can be written as HCx3dy +
y3dx. Let P(x, y) = x3and Q(x, y) = y3. Then, ∂Q
∂x = 0 and ∂P
∂y = 0. Hence,
the double integral reduces to zero.
Step 3: Since the double integral over the region enclosed by Cis zero, the
value of the line integral HC(y3dx +x3dy)is also zero.
Therefore, the value of the line integral HC(y3dx +x3dy)over the circle
x2+y2= 4 is 0.
Question 24
Question
Evaluate the line integral HC(x2+y2)dx+2xydy, where Cis the circle x2+y2= 4,
oriented counterclockwise.
Solution
Step 1: We will first rewrite the given line integral in terms of a potential
function. Let P(x, y) = x2+y2and Q(x, y) = 2xy. Then, we have ∂Q
∂x = 2y=
∂P
∂y , which implies that the integrand is exact. By Green’s Theorem, we have
IC
(x2+y2)dx + 2xydy =ZZD∂Q
∂x −∂P
∂y dA,
where Dis the region enclosed by C, which is the circle x2+y2= 4.
Step 2: The region Dcan be described using the polar coordinates x=
rcos(θ),y=rsin(θ), with 0≤r≤2and 0≤θ≤2π. Thus, dA =rdrdθ. The
integral becomes
ZZD∂Q
∂x −∂P
∂y dA =Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ.
Step 3: We can now evaluate the double integral:
Z2π
0Z2
0
(2rsin(θ)−2r)rdrdθ =Z2π
0Z2
0
2r2sin(θ)−2r2drdθ
=Z2π
0 2
3r3sin(θ)−r3r=2
r=0!dθ
=Z2π
016
3sin(θ)−8dθ
=−16
3cos(θ)−8θ2π
0
=−16
3(cos(2π)−cos(0)) −8(2π−0)
=−16
3−16π.
23
Therefore, the value of the line integral HC(x2+y2)dx + 2xydy around the
circle x2+y2= 4 oriented counterclockwise is −16
3−16π.
Question 25
Question
Let Cbe the curve defined by x2+y2= 4 oriented counterclockwise. Use
Green’s Theorem to evaluate the line integral
IC
(x2−y2)dx + 2xy dy.
Solution
Step 1: First, let’s calculate the partial derivatives of the given vector field
F(x, y) = (x2−y2,2xy).
∂
∂y (x2−y2) = −2y, ∂
∂x(2xy) = 2y.
Step 2: Now, Green’s Theorem states that for a simply connected region
Dbounded by a positively oriented, piecewise-smooth, simple closed curve C,
the line integral of a vector field
F= (P, Q)along Ccan be evaluated as
IC
F·dr =ZZD∂Q
∂x −∂P
∂y dA,
where Pand Qare the component functions of
F.
Step 3: Applying Green’s Theorem to the given line integral, we have
IC
(x2−y2)dx + 2xy dy =ZZD
(2y−(−2y)) dA = 4 ZZD
y dA.
Step 4: The region Denclosed by the curve Cis a circle of radius 2.
We can use polar coordinates to evaluate the double integral. The conversion
formula from Cartesian to polar coordinates is x=rcos θand y=rsin θ, with
dA =r dr dθ.
Step 5: Substituting y=rsin θand dA =r dr dθ into the double integral,
we get
4ZZD
y dA = 4 Z2π
0Z2
0
rsin θ·r dr dθ.
24
Step 6: Evaluating the double integral, we have
4Z2π
0Z2
0
rsin θ·r dr dθ = 4 Z2π
0r3cos θ
3
2
0sin θ dθ
= 4 Z2π
0
8
3cos θsin θ dθ
=32
3Z2π
0
1
2sin(2θ)dθ
=32
3−1
4cos(2θ)
2π
0
=32
3−1
4−(−1
4)
=32
3×0
= 0.
Therefore, the value of the line integral HC(x2−y2)dx + 2xy dy along the
curve Cis 0.
25
